Divide the following complex numbers. $\dfrac{-20+4i}{3+2i}$
Answer: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate, which is ${3-2i}$. $ \dfrac{-20+4i}{3+2i} = \dfrac{-20+4i}{3+2i} \cdot \dfrac{{3-2i}}{{3-2i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$. $ = \dfrac{(-20+4i) \cdot (3-2i)} {3^2 - (2i)^2} $ Evaluate the squares in the denominator and subtract them. $ = \dfrac{(-20+4i) \cdot (3-2i)} {(3)^2 - (2i)^2} $ $ = \dfrac{(-20+4i) \cdot (3-2i)} {9 + 4} $ $ = \dfrac{(-20+4i) \cdot (3-2i)} {13} $ The denominator now doesn't contain any imaginary unit multiples, so it is a real number. Note that when a complex number, $a + bi$ is multiplied by its conjugate, the product is always $a^2 + b^2$. Now, we can multiply out the two factors in the numerator. $ \dfrac{({-20+4i}) \cdot ({3-2i})} {13} $ $ = \dfrac{{-20} \cdot {3} + {4} \cdot {3 i} + {-20} \cdot {-2 i} + {4} \cdot {-2 i^2}} {13} $ $ = \dfrac{-60 + 12i + 40i - 8 i^2} {13} $ Finally, simplify the fraction. $ \dfrac{-60 + 12i + 40i + 8} {13} = \dfrac{-52 + 52i} {13} = -4+4i $